Question 21 and 22, Exercise 4.7

Solutions of Question 21 and 22 of Exercise 4.7 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Sum the series up to $n$ term: $1 \times 4+2 \times 7+3 \times 10+\cdots$

Solution.

Consider $T_k$ represents the $k$th term of the sereies, then $$\begin{align*}T_k&=k(3k+1) \\ &=3k^2+k. \end{align*}$$

Taking summation, we have $$\begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} (3k^2 +k)\\ & = 3 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \\ & = 3\left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2} \\ & = \frac{n(n+1)}{2}(2n+1+1) \\ & = \frac{n(n+1)}{2}(2n+2) \\ & = \frac{2n(n+1)}{2}(n+1) \\ & = n(n+1)^2 \end{align*}$$

Thus, the sum of the series is $\sum\limits_{k=1}^{n} T_{k} = n(n+1)^2.$

Sum the series up to $n$ term: $1 \times 3 \times 5+3 \times 5 \times 7+5 \times 7 \times 9+\cdots$ to $n$ term.

Solution.

Consider $T_k$ represents the $k$th term of the sereies, then $$\begin{align*}T_k&=(2k-1)(2k+1)(2k+3) \\ &=(4k^2-1)(2k+3)\\ &=8k^3+12k^2-2k-3\end{align*}$$

Taking summation, we have $$\begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} (8k^3+12k^2-2k-3)\\ & = 8 \sum_{k=1}^{n} k^3 +12 \sum_{k=1}^{n} k^2-2 \sum_{k=1}^{n} k -3 \sum_{k=1}^{n} 1 \\ & = 8\left(\frac{n(n+1)}{2}\right)^3+12\left(\frac{n(n+1)(2n+1)}{6}\right)-2\left( \frac{n(n+1)}{2} \right) - 3n \\ & = \left(n(n+1)\right)^3+2\left(n(n+1)(2n+1)\right)-\left( n(n+1) \right) - 3n \\ &=n[\left(n^2(n^3+3n^2+3n+1)\right)+2\left(2n^2+3n+1\right)-\left( (n+1) \right) - 3] \\ & = n[n^5+3n^4+3n^3+n^2+4n^2+6n+2-n-1 - 3] \\ & = n[n^5+3n^4+3n^3+5n^2+5n-2] \\ & = n(n+2)(n^4+n^3+n^2+3n-1) \end{align*}$$

Thus, the sum of the series is $\sum\limits_{k=1}^{n} T_{k} =n(n+2)(n^4+n^3+n^2+3n-1).$

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